Question: Let $h(x)=\dfrac{-4}{x^3}+\dfrac{1}{x}$. $h'(-2)=$
Explanation: The strategy We can first rewrite each rational term of $h$ as a negative power of $x$. Then, the derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Once we have $h'(x)$, we can plug $x=-2$ into it to find $h'(-2)$. Rewriting rational terms as negative powers $\begin{aligned} h(x)&=\dfrac{-4}{x^3}+\dfrac{1}{x} \\\\ &=-4x^{-3}+x^{-1} \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}h'(x) \\\\ &=\dfrac{d}{dx}(-4x^{-3}+x^{-1}) \\\\ &=-4\dfrac{d}{dx}(x^{-3})+\dfrac{d}{dx}(x^{-1}) \\\\ &=-4(-3x^{-4})+(-1)x^{-2} \\\\ &=12x^{-4}-1x^{-2} \\\\ &=\dfrac{12}{x^4}-\dfrac{1}{x^2} \end{aligned}$ Evaluating $h'(x)$ So we found that $h'(x)=\dfrac{12}{x^4}-\dfrac{1}{x^2}$. Let's plug $x=-2$ into $h'$ : $\begin{aligned} &\phantom{=}h'(-2) \\\\ &=\dfrac{12}{(-2)^4}-\dfrac{1}{(-2)^2} \\\\ &=\dfrac{12}{16}-\dfrac{1}{4} \\\\ &=\dfrac{3}{4}-\dfrac{1}{4} \\\\ &=\dfrac{2}{4} \\\\ &=\dfrac{1}{2} \end{aligned}$ In conclusion, $h'(-2)=\dfrac12$.